Q 8.81. - Elementary Statistics

Question

Medical Marijuana. Refer to Exercise $8.77$

a. The mean number of days that $30$ adolescents in substance abuse treatment used medical marijuana in the last $6$ months was $105.43$ Find a $95\%$ confidence interval for $\mu$ based on that data.

b. Compare the $95\%$ confidence intervals obtained here and in Exercise $8.77$ (a) by drawing a graph similar to Fig. $8.7$ on page $327$

c. Compare the margins of error for the two $95\%$ confidence intervals.

d. What principle is being illustrated?

Step-by-Step Solution

Answer

Part (a) $(93.979, 116.881)$

Part (b)  

Part (c) $5.726$

Part (d) The sample size is decreased and the confidence level is the same, which provides the increased margin of error.

Part (a) Step 1: Given information

$\overline{x}=105.43, n=30$ and $\sigma =32$

Part (a) Step 2: Concept

The formula used: Margin of error.

Part (a) Step 3: Calculation

Using MINITAB, calculate a $95\%$ confidence interval estimate of the population mean $\mu$

Consider $\overline{x}=105.43,n=30$and $\sigma =32$

Step 1: Select Stat $>$ Basic Statistics $>$ $1-$Sample $Z$ in MINITAB.

Step 2: In Summarized Data, enter $30$ as the sample size and $105.43$ as the mean.

Step 3: In the Standard deviation box, type $32$ for $s$

Step 4: Select Options and set the Confidence Level to $95$

Step 5: In the alternative, select not equal.

Step 6: In all dialogue boxes, click OK.

MINITAB output:

One-Sample Z

The assumed standard deviation $=32$

N	Mean	SE Mean	95% CI
30	105.430	5.842	(93.979, 116.881)

From the MINITAB output, the $95\%$ confidence interval estimate of the population mean $\mu$is $(93.979,116.881)$

Part (b) Step 1: Calculation

The confidence interval in Exercise $8.77$ is shown in the below graph:

The confidence interval in part (a) is shown in the below graph: 

Part (c) Step 1: Calculation

Obtain the margin of error for the $95\%$ confidence interval when $(93.979,116.881)$

The margin of error is,

$$\begin{gathered} \text{ Margin of error }=\frac{116.881-93.979}{2} \\ =\frac{22.902}{2} \\ =11.451 \end{gathered}$$

Thus, the margin of error for the $95\%$ confidence interval is $11.451$

From Exercise $8.77$, the $95\%$ confidence interval for $\mu$ is $(96.994,108.446)$

The margin of error is,

$$\begin{gathered} \text{ Margin of error }=\frac{108.446-96.994}{2} \\ =\frac{11.452}{2} \\ =5.726 \end{gathered}$$

Thus, the margin of error for the $95\%$ confidence interval is $5.726$

When compared to the margin of error obtained in Exercise $8.77$ this exercise has a bigger margin of error.

Part (d) Step 1: Explanation

The premise is that the sample size is reduced while the confidence level remains constant, resulting in a growing margin of error.